Observing an astronaut falling into a black hole

overlook1977

From what I understand, if an outside observer was to witness an astronaut falling into a black hole, relativity dictates that we would see him approach the black hole slower and slower until he ultimately "freezes" in our relative time due to time dilation. My question is, would we actually observe a frozen astronaut in space from our perspective (as in a photographic snapshot)? Doesn't the light become so redshifted that the astronaut's image would appear redder until ultimately fading away?

mathman

You are right about the light of his image being redshifted, even though he is frozen in space.

overlook1977

So my question is, would the light be so redshifted that we would not even be able to visibly see an image of the astronaut?

xantox

A finite number of photons is emitted before the astronaut crosses the horizon. In the continuum approximation, the light intensity decreases exponentially ∝ exp(-t/3√3 M) so that it will almost instantly completely disappear from sight.

MeJennifer

The number of photons emitted from the astronaut does not depend on how far he is away from the event horizon or even if he has crossed it or not.

pervect

Yes. According to "Black holes & Time warps" by Thorne, near the horizon there will be a characteristic doubling time for the wavelength of the emitted radiation (.14 milliseconds in the example Thorne discusses on pg 33 - in which an infalling observer shines out a coherent laser beam of a specified frequency).

Eventually there will be a last visible photon, as the laser beam quickly redshifts below the visible region.

In fact, given any non-zero frequency, there will be some time at which no photons are emitted above that frequency. In the book I mentioned, there will be a time of last emission of 4km radio waves, the lowest frequency that the monitoring observer can detect.

I believe this is also discussed in MTW's textbook "Gravitation", but I couldn't find the exact page. As Thorne is one of the authors of this book, I would expect that it would be the same answer as found in the more readilly accessible "Black Holes & Time Warps".

MeJennifer

Assuming the observer travels inertially, then it is the case that he measures the photons coming from the astronaut at decreasing frequencies and increasing intervals, but that does not imply that the photons were emitted at decreasing frequencies and increasing intervals.

Also, just because photons are no longer received by the observer does not mean they were not emitted by the astronaut, it just means that the spacetime paths of the photons no longer have an intersection point with the spacetime path of the observer, in fact after the astronaut passes the event horizon all the photon paths head for the singularity.

xantox

Regardless of that, he will only emit a finite number of photons before crossing the horizon. Very shortly after the observer will notice the exponential luminosity decay (or the exponential redshift, which is the same), the very last photon emitted before crossing the horizon will also have reached him.

Chris Hillman

This is a really bad way to think about it. See Geroch, General relativity from A to B (which has almost no mathematical prerequisites) and try to understand "gravitational redshifting" in terms of initially parallel radially outgoing null geodesics which diverge due to the curvature of spacetime.

No, and yes. If you have a suitable math/physics background (or even if not, since this next book also has superb pictures which basically tell the whole story), see the discussion in MTW, Gravitation in the section on an astronaut falling into a black hole :-/ (This is no doubt the section pervect was thinking of, and you won't find a more knowledgeable account of anything concerning gtr than you will find in MTW, with very infrequent allowances for the progress of physics since 1973.)

See http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html for the full bibliographic citations. Enjoy!